\section{1.12} 
\begin{frame}[allowframebreaks]{1.12. }

\vspace{-0.4cm}

1.12. Holonomic modules. 

Let $M \in \mu(\mathcal{D}_X)$. 

Then $M$ is holonomic if it is coherent and if $\dim_x \mathrm{SS}(M) \le \dim X$ for all $x \in X$.

Note that if the inequality is strict, then $M_x = 0$ in view of 1.10(ii). 

We have again:

Let $M \in \mathrm{coh}(\mathcal{D}_X)$ and $x \in X$. 

If $M_x \ne 0$, then $M$ is holonomic around $x$ if and only if $\mathrm{Ext}^j_{\mathcal{D}_{X,x}}(M, \mathcal{D}_X) = 0$ for $j \ne \dim X$.

As in the case of the Weyl algebra. (V,1.11) we see that if
\[
0 \to M' \to M \to M'' \to 0
\]
is an exact sequence of $\mathcal{D}_X$-modules, then $M$ is holonomic if and only if $M'$ and $M''$ are so. 

The holonomic $\mathcal{D}_X$-modules form a full subcategory of $\mathrm{coh}(\mathcal{D}_X)$, to be denoted $\mathrm{hol}(\mathcal{D}_X)$. 

Also, 1.9 implies that $M$ is holonomic if and only if $\mathrm{Ext}^i_{\mathcal{D}_X}(M, \mathcal{D}_X) = 0$ for $i \ne \dim X$.

We also note that (V,1.16.1) implies that any holonomic module has finite length. 

It is enough to show that each point $x \in X$ has some affine neighborhood $U$ in which this is the case. 

Assume $U,V = \mathbb{A}^n - S$ and $f$ to be as in the proof of 1.10. 

Let $M$ be a holonomic $\mathcal{D}(U)$-module. 

It is then a $\mathcal{D}(V)$-module, which is holonomic by 1.10(2). 

We have $\mathbb{A}_n \subset \mathcal{D}(V)$ and $M$ is holonomic over $\mathbb{A}_n$ by (V,1.16.3), hence has finite length over $\mathbb{A}_n$ (V,1.16.1). 

Then $M = M[f^{-1}]$ has also finite length over $\mathcal{D}(V) = \mathbb{A}_n[f^{-1}]$.

\end{frame}

